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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution:</dfn> <span class="process-math">\(f(x, y)=y^2\)</span> and <span class="process-math">\(\frac{\partial f}{\partial y}=2 y\text{.}\)</span> Thus, according to the theorem, there is a unique solution which is valid in some interval containing <span class="process-math">\(x=0\text{.}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
\begin{aligned}
&amp;\frac{\textrm{d} y}{\textrm{d} x}=y^2~\rightarrow~\frac{1}{y^2} \textrm{d} y=\textrm{d} x,\\
&amp;-\frac{1}{y}=x+C~\rightarrow~y=-\frac{1}{x+C}.
\end{aligned}\tag{2.4.2}
\end{equation}
</div>
<p class="continuation">Then using the initial value condition <span class="process-math">\(y(0)=1\text{,}\)</span> one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
1=-\frac{1}{C}~\rightarrow~C=-1.
\end{equation*}
</div>
<p class="continuation">Thus, the solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=\frac{1}{1-x}.
\end{equation*}
</div>
<p class="continuation">We require that <span class="process-math">\(x \neq 1\text{,}\)</span> i. e., <span class="process-math">\(-\infty &lt; x &lt; 1\)</span> is the interval in which the solution is valid. If we take <span class="process-math">\(y(0)=-1\text{,}\)</span> then one has <span class="process-math">\(C=1\)</span> and the solution to the initial value problem is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=-\frac{1}{1+x},
\end{equation*}
</div>
<p class="continuation">and the valid interval is then <span class="process-math">\((-1, \infty)\text{.}\)</span> This example shows that the valid intervals depend on the initial conditions.</p>
<span class="incontext"><a href="sec2_4.html#p-34" class="internal">in-context</a></span>
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